I am referring to molecular orbital (MO) theory. In the nitrogen molecule (and elements of lower atomic number) $\textrm{sp}$ mixing occurs and the sigma orbital set is raised above the pi orbital set. Shouldn't that mean that pi bond is stronger than sigma bond in this case?


2 Answers 2


You can actually calculate orbital "energies" with MO theory. The resulting "energies" for $\ce{N2}$ are the following (calculated with molpro and CASSCF(10,8)/aug-cc-pVTZ):

Orbital "energies" in atomic units

orbital energy/a.u.
g -1.11
g -0.99
u -0.77
πu -0.59
πg 0.29
u 1.22

As you can see, the "energy" of the π orbitals is above the "energy" of the 2σg, which is even below the 1σu "energy". This disagrees with virtually every MO scheme of $\ce{N2}$ you can find. So why is that?

In the end, MO schemes are used in what is called "qualitative molecular orbital theory" (QMO theory). There are usually no underlying calculations (or the calculations were performed in the 60s) and the schemes are adapted in order to explain and understand the stability and reactivity of molecules.

On the other hand, there is computational MO theory, which could in principle be used to improve these diagrams. Why is this not done? In the end, electronic states of a molecule exist and have energies, while orbitals don't. They are not in any way 'real' or observable and are in general best understood as a tool for building approximate wave functions. The orbital "energies" are eigenvalues of the so-called Fock operator, but they should not be given too much attention. For the most accurate wave functions (the so-called full-CI wave functions), the orbital energies are actually arbitrary.

In the end, MO schemes are a tool used to understand and simplify the discussion of chemical bonding. If they cease to be simple, they should not be used.

As for the discussion of the π bond energy vs. the σ bond energy: again, there are not really separate π and σ bonds. However, if you want to calculate some energies aligning with these concepts, valence bond theory is better at that than MO theory.

  • 1
    $\begingroup$ On Chemistry mathematical and chemical expressions can be formatted using MathJax (and LaTeX Syntax). If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details. $\endgroup$ Commented Jul 29, 2021 at 17:55
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    $\begingroup$ MO schemes are essentially a visualisation of a single determinante self-consistent field, or Hartree-Fock. Sorry for being nitpicky here, but the Fock-Operator, whether it is so-called or not, doesn't actually have eigenvalues as it is dependent on its own solutions. This obviously didn't stop generations of chemists to use it as such. Are for a FCI wave function the "energies" really arbitrary, or are the not defined? But I actually like where this answer is heading to. $\endgroup$ Commented Jul 29, 2021 at 18:11
  • $\begingroup$ @Martin-マーチン thanks for the syntax improvements! Regarding your other comment: you are right, that MO schemes are essentially a visualization of a single determinant, I just performed the CASSCF calculation instead of HF in order to have reasonable energies for the unoccupied orbitals, which are not optimized with HF. The Fock operator really has these eigenvalues: see here. $\endgroup$
    – Libavi
    Commented Jul 30, 2021 at 8:56
  • $\begingroup$ As for the FCI wave function: with FCI, you build all possible configurations with all orbitals, construct the wave function as a linear combination of those and optimize only the linear coefficients. The shape of the orbitals does not matter, if they span the same vector space. They can be constructed by any orthogonalization procedure performed on the atomic basis functions. $\endgroup$
    – Libavi
    Commented Jul 30, 2021 at 9:00
  • $\begingroup$ Not the first time Wikipedia is wrong. The equation they mention is a pseudo-eigenvalue equation, you'll see that when you derive it. I guess I skipped the important steps here: LCAO (Linear Combination of Atomic Orbitals) and Phases $\endgroup$ Commented Jul 30, 2021 at 18:18

There is a similar question on StackExchange from 6 years ago (Ref 1), from which I have taken Diagram #1:


enter image description here

And a quote from the answer: "Many of the diagrams (on the internet) are different one from another - that's a tip off that s-p mixing is not something that can always be predicted in advance." Something like Diagram #2 below, which implies a very different degree of hybridization (none). Just to be clear, "N3" means the nitrogen molecule (which is N2).


enter image description here

The four orbitals (2 sigma atomic and 2 pi atomic) interact and separate into 4 sigma bonds, two bonding and two antibonding. The lowest three sigma MOs are filled in N2. The take-away from both diagrams is that the three lowest energy sigma MOs are energetically more stable than the atomic s and p orbitals, even tho formally, the uppermost filled sigma level is antibonding.

The degree of splitting is dependent on the amount of hybridization entered into the calculation, and if that is small, the actual energy levels will resemble Diagram #2, and if the amount of hybridization calculated is large, the molecular orbital energies calculated will resemble Diagram #1. Calculations are an important part of the discussion.

Both diagrams predict an increase in ionization potential on going from atomic N to molecular N2 (actual ionization potential for N is 1402 kJ/mol; for N2 is 1503 kJ/mol), but the difference does not allow a definitive choice between the models.

So it comes down to estimating the hybridization and doing a calculation. I suspect that if you try to determine something about the first ionized state of N2, it would resemble Diagram #1 more than #2, because the nitrogen atoms would be drawn closer together.

Another differentiation would be to see how nitrogen molecules bind to other atoms. Ref 2 indicates that the bonding is linear, in agreement with both diagrams. Nitrogen sometimes (rarely) bonds with two atoms (a side bond). The question allows for a lot of pictorial discussion, but not much in a numerical sense.

It could be looked at as a comparison between a stabilizing, but formally sigma antibond against a pi bond and an interesting way to bring up hybridization and overlaps.

enter image description here

Ref 1. In molecular orbital theory, why does s-p mixing in the dinitrogen molecule not effect the 1σᵤ orbital?

Ref 2. https://en.wikipedia.org/wiki/Transition_metal_dinitrogen_complex

  • $\begingroup$ This is kind of in a tangent to the question OP asked. Also, I would question that the two MO diagrams you drew above are actually different. They have the same ordering of the orbitals, just the dotted lines joining AO and MO are different. $\endgroup$
    – S R Maiti
    Commented Jul 25, 2021 at 9:36
  • $\begingroup$ @S R Maiti: The 2nd says that the N s orbitals overlap (sigma bond), and three p orbitals form a sigma and two pi bonds. The first says an s and a p orbital on each N hybridize to sp orbitals: two lobed, one lobe much larger. That's the 4 sigma MOs in the third diagram. (The pi double bonds in N2 are the same in both diagrams.) Draw the 4 orbital shapes with a) no hybridization and b) with complete hybridization. Is it "fair" to compare a filled sigma anti-bond (#3) with a pi double bond? I don't think so. $\endgroup$ Commented Jul 25, 2021 at 12:38
  • $\begingroup$ Oh, so you are referring to the third diagram as the 2nd? I thought you meant the two diagrams at the top. It's a bit confusing to follow. $\endgroup$
    – S R Maiti
    Commented Jul 25, 2021 at 15:00
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    $\begingroup$ Hybridisation and s-p-mixing are completely different things! You can always have an explanation with hybrid orbitals as these are just mathematical descriptions. It will tell you absolutely nothing about the order of the molecular orbitals. And on that note, the diagram 2 is almost completely wrong. And moving further, it is impossible to conclude on bond strengths from an MO diagram. On the other hand, you can calculate (predict) bond strengths (hypothetically of course) without ever needing to resort to hybridisation or drawing a MO diagram. $\endgroup$ Commented Jul 25, 2021 at 18:57
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    $\begingroup$ No, on all accounts. First, I've made my objections to Ron's answer already there. Second, the second MO diagram is not inaccurate, it's wrong. If it were assuming no s-p-mixing, then the second sigma(g) orbital must be lower in energy than the pi orbitals. Third, and I have no way of saying that differently, it's impossible to conclude to bond strength from a MO diagram. The orbital energies do not represent bond energies, as they also do not add up to the total energy of the wave function. A MO diagram cannot capture electron correlation, but that's crucial for bonding. $\endgroup$ Commented Jul 26, 2021 at 17:18

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